Number Series Questions for IBPS PO Exam

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Number Series Questions for IBPS PO Exam

Download top-20 Number Series Questions PDF based on previously asked questions in IBPS PO exams. Go through important Number Series Questions for IBPS PO Bank prelims and mains exam.

Instructions

In each of these questions a number series is given. In each series â€˜â€™only oneâ€™â€™ number is wrong. Find out the â€˜â€™wrong numberâ€™â€™

Question 1:Â 2, 8, 12, 20, 30, 42, 56

a)Â 8

b)Â 42

c)Â 30

d)Â 20

e)Â 12

Question 2:Â 7, 13, 25, 49, 97, 194, 385

a)Â 13

b)Â 49

c)Â 97

d)Â 194

e)Â 25

Instructions

What should come in place of question mark (?) in the following numbers series

Question 3:Â 22, 23, 27, 36, 52, 77, ?

a)Â 111

b)Â 109

c)Â 113

d)Â 117

e)Â 115

Instructions

What should come in place of the question mark (?) in the following numbers series ?

Question 4:Â 2 16 112 672 3360 13440 ?

a)Â 3430

b)Â 3340

c)Â 40320

d)Â 43240

e)Â None of these

Question 5:Â 4 9 19 ? 79 159 319

a)Â 59

b)Â 39

c)Â 49

d)Â 29

e)Â None of these

Question 6:Â 4000 2000 1000 500 250 125 ?

a)Â 80

b)Â 65

c)Â 62.5

d)Â 83.5

e)Â None of these

Question 7:Â 588 563 540 519 ? 483 468

a)Â 500

b)Â 496

c)Â 494

d)Â 490

e)Â None of these

Question 8:Â 121 ? 81 64 49 36 25

a)Â 92

b)Â 114

c)Â 98

d)Â 100

e)Â None of these

Instructions

Find the next term in the series.

Question 9:Â 442, 378, 346, 330, 322

a)Â 320

b)Â 318

c)Â 310

d)Â 314

e)Â 316

Question 10:Â 20,40,61,84,111

a)Â 138

b)Â 146

c)Â 136

d)Â 144

e)Â None of these

Question 11:Â 1,7,17,31,49,71

a)Â 91

b)Â 86

c)Â 97

d)Â 87

e)Â None of these

Question 12:Â 2,4,8,14,22,32

a)Â 46

b)Â 40

c)Â 42

d)Â 44

e)Â None of these

Question 13:Â 2,12,36,80,150

a)Â 252

b)Â 246

c)Â 254

d)Â 244

e)Â None of these

Instructions

In each of the following number series, a wrong number is given. FInd out the wrong number.

Question 14:Â 2, 3, 6, 18, 109,Â 1944, 209952

a)Â 3

b)Â 6

c)Â 18

d)Â 109

e)Â 1944

Question 15:Â 1 3 6 11 20 39 70

a)Â 3

b)Â 39

c)Â 11

d)Â 20

e)Â 6

Question 16:Â 2 13 27 113 561 3369 23581

a)Â 13

b)Â 27

c)Â 113

d)Â 561

e)Â 3369

Question 17:Â 50 51 47 56 42 65 29

a)Â 51

b)Â 47

c)Â 56

d)Â 42

e)Â 65

Question 18:Â 3 9 23 99 479 2881 20159

a)Â 9

b)Â 23

c)Â 99

d)Â 479

e)Â 2881

Question 19:Â 2 4 5 8 13 21 34

a)Â 4

b)Â 5

c)Â 8

d)Â 13

e)Â 21

Instructions

In each of these questions a number series is given. In each series â€˜â€™only oneâ€™â€™ number is wrong. Find out the â€˜â€™wrong numberâ€™â€™

Question 20:Â 4, 12, 42, 196, 1005, 6066, 42511

a)Â 12

b)Â 42

c)Â 1005

d)Â 196

e)Â 6066

Instead of 8, it should be 6. Then the difference the consecutive terms will be in AP.

6-2 = 4

12-6 = 6

20-12 = 8

30-20 = 10

42-30 = 12

56-42 = 14

13 = 7+6

25 = 13+12

49 = 25+24

97 = 49+48

193Â = 97+96

385 = 193+192

So 194 is an odd number out.

The difference between consecutive terms of the series is a perfect square.

For example,

23-22 = 1

27-23 = 4

36-27 = 9

52-36 = 16

77-52 = 25

Hence, the next difference between the numbers will be 36 and the required number is 77+36 = 113

Here, in the given series is obtained by nth term = previous term * (10 – n)

We are supposed to calculate 7th term = 13440 * (10 -7) = 13440 * 3 = 40320

Therefore, option C is correct.

In the bgiven series, the nth term is obtained as,
nth term = previous term + 5 * 2^(n-2)
Therefore, 4th term = 19 + 5 * 2^(2) =39
Hence, the correct option is option B.

The given series is geometric progression with common ratio of 0.5.

Therefore, 7th term = 6th term * 0.5 = 125 *0.5 = 62.5

Hence, the correct answer is option C.

588 563 540 519 ? 483 468
588 – 563 = 25
563 – 540 = 23
540 – 519 =21
The difference is getting reduced by 2
Therefore, 519 – x = 19
x = 500
The correct option is option A.

Here, 81 64 49 36 25
81 – 64 = 17
64 – 49 = 15
49 – 36 = 13
36 – 25 = 11

Therefore, the difference is in arithmetic progression with common difference of 2.
Hence,
x – 81 = 19
x = 100
Therefore, the correct option is option D.

In the given series the sum of two consecutive terms are in GP
442-378 = 64
378-346 = 32
346-330 = 16
330-322 = 8
322-x = 4
x = 318

20 = 19+1
40 = (19*2)+2
61 = (19*3)+4
84 = (19*4)+8
111 = (19*5)+16
Next term = (19*6)+32 = 146

1 =$2(1)^2-1$
7 =$2(2)^2-1$
17 =$2(3)^2-1$
31 =$2(4)^2-1$
49 =$2(5)^2-1$
71 =$2(6)^2-1$
Next terms =$2(7)^2-1 = 97$

4-2 = 2
8-4 = 4
14-8 = 6
22-14 = 8
32-22 = 10
Next term = 32+12 = 44

2 =$n^3+n^2$
12 =$2^3+2^2$
36 =$3^3+3^2$
80 =$4^3+4^2$
150 =$5^3+5^2$
Next term =$6^3+6^2 = 252$

Each number in the series given is the product of the earlier two numbers.

For example, 2*3 = 6
3*6 = 18
6*18 = 108
18*108 = 1944
1944*108 = 209952

In this case, the number which is wrong in the series is 109. It should be replaced by 108.

$1 \times 2+1 = 3$
$3 \times 2+0 = 6$
$6 \times 2-1 = 11$
$11 \times 2-2 = 20$
$20 \times 2-3 = 37$
$37 \times 2-4 = 70$
Hence, 39 is the wrong number here.

$(2\times2)+7 = 11$
$(11\times3)-6 = 27$
$(27\times4)+5 = 113$
$(113\times5)-4 = 561$
$(561\times6)+3 = 3369$
$(3369\times7)-2 = 23581$

13 is the wrong number.

$50+1^2 = 51$
$51-2^2 = 47$
$47+3^2 = 56$
$56-4^2 = 40$
$40+5^2 = 65$
$65-6^2 = 29$

42 is the incorrect number

$(3\times2)+3 = 9$
$(9\times3)-4 = 23$
$(23\times4)+5 = 97$
$(97\times5)-6 = 479$
$(479\times6)+7 = 2881$
$(2881\times7)-8 = 20159$
Hence, it should be 97 instead of 99.

2 + 1 = 3
3 + 2 = 5
5 + 3 = 8
8 + 5 = 13
13 + 8 = 21
21 + 13 = 34

4 is the incorrect number

$T_n = T_{n-1}*n + n^2$

12 = $4*2 + 2^2$

45 = $12*3 + 3^2$

196 = $45*4+4^2$

and so on.

So 42 is the wrong term in the series.

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